Cho hàm số f(x)= căn bậc hai (x^2+2x+4)-căn bậc hai (x^2-2x+4). Khẳng định nào sau đây đúng?

* Hướng dẫn giải

$f\left(x\right)=\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}$

Ta có:

$\underset{x\to +\infty }{lim}f\left(x\right)=\underset{x\to +\infty }{lim}\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)$

$=\underset{x\to +\infty }{lim}\frac{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}$

$=\underset{x\to +\infty }{lim}\frac{(x^2+2x+4)-(x^2-2x+4)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}$

$\begin{array}{l}=\underset{x\to +\infty }{lim}\frac{4x}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to +\infty }{lim}\frac{4}{\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=2\end{array}$

$\begin{array}{l}\underset{x\to -\infty }{lim}f\left(x\right)=\underset{x\to -\infty }{lim}\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\\ =\underset{x\to -\infty }{lim}\frac{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}\\ =\underset{x\to -\infty }{lim}\frac{(x^2+2x+4)-(x^2-2x+4)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{lim}\frac{4x}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{lim}\frac{\frac{4x}{x}}{\frac{\sqrt{x^2+2x+4}}{x}+\frac{\sqrt{x^2-2x+4}}{x}}\\ =\underset{x\to -\infty }{lim}\frac{4}{\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=\frac{4}{-1-1}=-2\end{array}$

$\Rightarrow \underset{x\to +\infty }{\lim }f\left(x\right)=-\underset{x\to -\infty }{\lim }f\left(x\right)$